Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $a = \dfrac{-8t - 40}{t^2 - 6t - 40} \times \dfrac{t^2 - 10t}{-7t - 35} $
Explanation: First factor the quadratic. $a = \dfrac{-8t - 40}{(t - 10)(t + 4)} \times \dfrac{t^2 - 10t}{-7t - 35} $ Then factor out any other terms. $a = \dfrac{-8(t + 5)}{(t - 10)(t + 4)} \times \dfrac{t(t - 10)}{-7(t + 5)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ -8(t + 5) \times t(t - 10) } { (t - 10)(t + 4) \times -7(t + 5) } $ $a = \dfrac{ -8t(t + 5)(t - 10)}{ -7(t - 10)(t + 4)(t + 5)} $ Notice that $(t + 5)$ and $(t - 10)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ -8t(t + 5)\cancel{(t - 10)}}{ -7\cancel{(t - 10)}(t + 4)(t + 5)} $ We are dividing by $t - 10$ , so $t - 10 \neq 0$ Therefore, $t \neq 10$ $a = \dfrac{ -8t\cancel{(t + 5)}\cancel{(t - 10)}}{ -7\cancel{(t - 10)}(t + 4)\cancel{(t + 5)}} $ We are dividing by $t + 5$ , so $t + 5 \neq 0$ Therefore, $t \neq -5$ $a = \dfrac{-8t}{-7(t + 4)} $ $a = \dfrac{8t}{7(t + 4)} ; \space t \neq 10 ; \space t \neq -5 $